Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

HOLOOLY
TEXTBOOKS

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

HOLOOLY
HELP DESK

Need Help? We got you covered.

Q. 14.2

Use the mass of the $W ^{-}$ particle to estimate the range of the weak interaction responsible for the neutron beta decay.

Strategy This calculation is much like the one we did in Example 14.1, when we discussed Yukawa’s meson. In that case we knew the nuclear force range, but in this case, the force range is unknown. We use Equation (14.4) to find the force range $R_{ W }$.

$R=\frac{\hbar}{2 m c}=\frac{\hbar c}{2 m c^{2}}$ (14.4)

Verified Solution

The range $R_{ W }$ becomes

\begin{aligned}R_{ W } &=\frac{\hbar c}{2 m_{ W } c^{2}} \\&=\frac{1.973 \times 10^{2} eV \cdot nm }{2\left(80.4 GeV / c^{2}\right)\left(c^{2}\right)}=1.2 \times 10^{-18} m\end{aligned}

In this case, it may not be true that the $W ^{-}$ travels near the speed of light, because it is such a massive particle. The calculation here is therefore an upper limit. Note that in this case the violation of the conservation of energy is extreme, so the violation must occur over a very short period of time. We calculate $\Delta t$ from Equation (14.2) to be

$\Delta t \approx \frac{\hbar}{2 \Delta E}=\frac{\hbar}{2 m_{\pi} c^{2}}$ (14.2)

\begin{aligned}\Delta t &=\frac{\hbar}{2 m_{ W } c^{2}} \\\Delta t &=\frac{1.055 \times 10^{-34} J \cdot s }{2(80.4 GeV )}\left(\frac{1 GeV }{1.6 \times 10^{-10} J }\right) \approx 4 \times 10^{-27} s\end{aligned}

The lifetime of the neutron is much longer than the time it takes for the decay process itself.